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"question": {
"questionId": "2625",
"questionFrontendId": "2536",
"categoryTitle": "Algorithms",
"boundTopicId": 2057619,
"title": "Increment Submatrices by One",
"titleSlug": "increment-submatrices-by-one",
"content": "<p>You are given a positive integer <code>n</code>, indicating that we initially have an <code>n x n</code>&nbsp;<strong>0-indexed</strong> integer matrix <code>mat</code> filled with zeroes.</p>\n\n<p>You are also given a 2D integer array <code>query</code>. For each <code>query[i] = [row1<sub>i</sub>, col1<sub>i</sub>, row2<sub>i</sub>, col2<sub>i</sub>]</code>, you should do the following operation:</p>\n\n<ul>\n\t<li>Add <code>1</code> to <strong>every element</strong> in the submatrix with the <strong>top left</strong> corner <code>(row1<sub>i</sub>, col1<sub>i</sub>)</code> and the <strong>bottom right</strong> corner <code>(row2<sub>i</sub>, col2<sub>i</sub>)</code>. That is, add <code>1</code> to <code>mat[x][y]</code> for all <code>row1<sub>i</sub> &lt;= x &lt;= row2<sub>i</sub></code> and <code>col1<sub>i</sub> &lt;= y &lt;= col2<sub>i</sub></code>.</li>\n</ul>\n\n<p>Return<em> the matrix</em> <code>mat</code><em> after performing every query.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/11/24/p2example11.png\" style=\"width: 531px; height: 121px;\" />\n<pre>\n<strong>Input:</strong> n = 3, queries = [[1,1,2,2],[0,0,1,1]]\n<strong>Output:</strong> [[1,1,0],[1,2,1],[0,1,1]]\n<strong>Explanation:</strong> The diagram above shows the initial matrix, the matrix after the first query, and the matrix after the second query.\n- In the first query, we add 1 to every element in the submatrix with the top left corner (1, 1) and bottom right corner (2, 2).\n- In the second query, we add 1 to every element in the submatrix with the top left corner (0, 0) and bottom right corner (1, 1).\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/11/24/p2example22.png\" style=\"width: 261px; height: 82px;\" />\n<pre>\n<strong>Input:</strong> n = 2, queries = [[0,0,1,1]]\n<strong>Output:</strong> [[1,1],[1,1]]\n<strong>Explanation:</strong> The diagram above shows the initial matrix and the matrix after the first query.\n- In the first query we add 1 to every element in the matrix.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 500</code></li>\n\t<li><code>1 &lt;= queries.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= row1<sub>i</sub> &lt;= row2<sub>i</sub> &lt; n</code></li>\n\t<li><code>0 &lt;= col1<sub>i</sub> &lt;= col2<sub>i</sub> &lt; n</code></li>\n</ul>\n",
"translatedTitle": "子矩阵元素加 1",
"translatedContent": "<p>给你一个正整数 <code>n</code> ,表示最初有一个 <code>n x n</code> 、下标从 <strong>0</strong> 开始的整数矩阵 <code>mat</code> ,矩阵中填满了 0 。</p>\n\n<p>另给你一个二维整数数组 <code>query</code> 。针对每个查询 <code>query[i] = [row1<sub>i</sub>, col1<sub>i</sub>, row2<sub>i</sub>, col2<sub>i</sub>]</code> ,请你执行下述操作:</p>\n\n<ul>\n\t<li>找出 <strong>左上角</strong> 为 <code>(row1<sub>i</sub>, col1<sub>i</sub>)</code> 且 <strong>右下角</strong> 为 <code>(row2<sub>i</sub>, col2<sub>i</sub>)</code> 的子矩阵,将子矩阵中的 <strong>每个元素</strong> 加 <code>1</code> 。也就是给所有满足 <code>row1<sub>i</sub> &lt;= x &lt;= row2<sub>i</sub></code> 和 <code>col1<sub>i</sub> &lt;= y &lt;= col2<sub>i</sub></code> 的 <code>mat[x][y]</code> 加 <code>1</code> 。</li>\n</ul>\n\n<p>返回执行完所有操作后得到的矩阵 <code>mat</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/11/24/p2example11.png\" style=\"width: 531px; height: 121px;\" /></p>\n\n<pre>\n<strong>输入:</strong>n = 3, queries = [[1,1,2,2],[0,0,1,1]]\n<strong>输出:</strong>[[1,1,0],[1,2,1],[0,1,1]]\n<strong>解释:</strong>上图所展示的分别是:初始矩阵、执行完第一个操作后的矩阵、执行完第二个操作后的矩阵。\n- 第一个操作:将左上角为 (1, 1) 且右下角为 (2, 2) 的子矩阵中的每个元素加 1 。 \n- 第二个操作:将左上角为 (0, 0) 且右下角为 (1, 1) 的子矩阵中的每个元素加 1 。 \n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/11/24/p2example22.png\" style=\"width: 261px; height: 82px;\" /></p>\n\n<pre>\n<strong>输入:</strong>n = 2, queries = [[0,0,1,1]]\n<strong>输出:</strong>[[1,1],[1,1]]\n<strong>解释:</strong>上图所展示的分别是:初始矩阵、执行完第一个操作后的矩阵。 \n- 第一个操作:将矩阵中的每个元素加 1 。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 500</code></li>\n\t<li><code>1 &lt;= queries.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= row1<sub>i</sub> &lt;= row2<sub>i</sub> &lt; n</code></li>\n\t<li><code>0 &lt;= col1<sub>i</sub> &lt;= col2<sub>i</sub> &lt; n</code></li>\n</ul>\n",
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"Imagine each row as a separate array. Instead of updating the whole submatrix together, we can use prefix sum to update each row separately.",
"For each query, iterate over the rows i in the range [row1, row2] and add 1 to prefix sum S[i][col1], and subtract 1 from S[i][col2 + 1].",
"After doing this operation for all the queries, update each row separately with S[i][j] = S[i][j] + S[i][j - 1]."
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